Problem set 7 1) Assume that the flowrate, Q, of a gas from a smokestack is a function of the density of the ambient air, ρa, the density of the gas, ρg, within the stack, the acceleration of gravity, g, and the height and diameter of the stack, h and d, respectively. Use ρa, d, and g as repeating variables to develop a set of pi terms that could be used to describe this problem. g h Q = φ , 1/ 2 d g a d 5/ 2 2) The velocity, V, of a spherical particle falling slowly in a viscous liquid can be expressed as V = f(d, μ, γ, γs) where d is the particle diameter, μ the liquid viscosity, and γ and γs the specific weight of the liquid and particle, respectively. Develop a set of dimensionless parameters that can be used to investigate this problem. Vμ = d 2γ Therefore γ f s γ 3) A cylinder with a diameter D floats upright in a liquid as shown in Figure. When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequency, ω. Assume that this frequency is a function of the diameter, D, the mass of the cylinder, m, and the specific weight, γ, of the liquid. Determine, with the aid of dimensional analysis, how the frequency is related to these variables. ω = CD m 4) The water flowrate, Q, in an open rectangular channel can be measured by placing a plate across the channel as shown in the Figure shown. This type of a device is called a weir. The height of the water, H, above the weir crest is referred to as the head and can be used to determine the flowrate through the channel. Assume that Q is a function of the head, H, the channel width, b, and the acceleration of gravity, g. Determine a suitable set of dimensionless variables for this problem. Q b = f 1/ 2 H g H 5/ 2 5) A thin flat plate having a diameter of 0.2 ft is towed through a tank of oil (γ = 53 lb/ft3) at a velocity of 6 ft/s. The plane of the plate is perpendicular to the direction of motion, and the plate is submerged so that wave action is negligible. Under these conditions the drag on the plate is 1.5 lb. If viscous effects are neglected, predict the drag on a geometrically similar, 2.4-ft-diameter plate that is towed with a velocity of 4 ft/s through water at 60 °F under conditions similar to those for the smaller plate. slug 1.94 3 2 2 ρ V d ft D= Dm = slug ρm Vm2 d m2 1.645 3 ft 2 ft 4 2.4 ft 2 s 1.5lb f = 113 lb f 2 2 ft .2 ft 6 s 6) The drag on a small, completely submerged solid body having a characteristic length of 2.1 mm and moving with a velocity of 10 m/s through water is to be determined with the aid of a model. The length scale is to be 60, which indicates that the model is to be larger than the prototype. Investigate the possibility of using either an unpressurized wind tunnel or a water tunnel for this study. Determine (a) the required velocity in the wind tunnel, (b) the required velocity in the water tunnel. m2 s m2 vwater = 1.12 106 s vair = 1.46 105 (a) Therefore for air, unpressurized wind tunnel velocity would be m2 s 1 10 m = 2.17 m Vm = m 2 60 s s 1.12×10-6 s 1.46×10-5 (b) For water m2 s 1 10 m = 2.17 m = .167 m Vm = m2 60 s s s 1.12×10-6 s 1.12×10-6 7) Consider a category five hurricane that has a maximum wind speed of 160 mph at the eye wall, 10 miles from the center of the hurricane. If the flow in the hurricane outside of the hurricane's eye is approximated as a free vortex, determine the wind speed at location (a) 50 mi, (b) 40 mi and (c) 30 mi from the center of the storm. (a) vθ = (b) vθ = 160 mph 10 mi =32 mph 50 mi 160 mph 10 mi =40 mph 40 mi (c) vθ = 160 mph 10 mi =53.33 mph 30 mi 8) A massive, precisely machined, 6-ft-diameter granite sphere rests upon a 4-ft diameter cylindrical pedestal as shown in the Figure. When the pump is turned on and the water pressure within the pedestal reaches 8 psi, the sphere rises off the pedestal, creating a 0.003-in. gap through which the water flows. The sphere can then be rotated about any axis with minimal friction. Estimate the pump flowrate, Q0, required to accomplish this. Assume the flow in the gap between the sphere and the pedestal is essentially viscous flow between fixed, parallel plates. The volumetric flow rate per unit width, q, for steady laminar flow between fixed plates with gap h and length l is given by q = 2h3 ΔP 3μl 3 3 13π -4 ft -3 ft Q0 = ft 1.92×10 =2.617×10 3 s s